Explanation

(a)
Given, $\sin{A - \cos{B = \cos C}}$
$\Rightarrow \ \sin{A = \cos B + \cos C}$
$\Rightarrow \ 2\sin{\frac{A}{2}\cos{\frac{A}{2} = 2\cos\left( \frac{B + C}{2} \right)}}\cos\left( \frac{B - C}{2} \right)$
$\Rightarrow \ 2\sin{\frac{A}{2}\cos{\frac{A}{2} = 2\sin{\frac{A}{2}\cos\left( \frac{B - C}{2} \right)}}}$
$\Rightarrow \ \cos{\frac{A}{2} = \cos{\left( \frac{B - C}{2} \right)\ \ \ \ \left\lbrack \because\ \ \sin{\left( \frac{A}{2} \right) \neq 0} \right\rbrack}}$
$\Rightarrow \ \frac{A}{2} = \frac{B - C}{2}\ \ \Rightarrow \ \ \ A = B - C$
But $A + B + C = \pi$, therefore
$2B = \pi$
$\Rightarrow \ \ B = \frac{\pi}{2}$